Question 16.1: A 2-m-long pin-ended column of square cross section is to be...

(a) For the 100-kN Load. Using the given factor of safety, we make

$P_{cr} = 2.5(100  kN) = 250  kN               L = 2  m                E = 13  GPa$

in Euler’s formula (16.11) and solve for I. We have

$P_{cr} =\frac{π^{2}EI}{L^{2}}$                         (16.11)

$I =\frac{P_{cr}L^{2}}{π^{2}E} = \frac{(250 × 10^{3}  N) (2  m)^{2}}{π^{2} (13 × 10^{9}  Pa)} = 7.794 × 10^{-6}  m^{4}$

Recalling that, for a square of side a, we have $I = a^{4}/12$, we write

$\frac{a^{4}}{12} = 7.794 × 10^{-6}  m^{4}                 a = 98.3   mm ≈ 100  mm$

We check the value of the normal stress in the column:

$σ = \frac{P}{A} = \frac{100  kN}{(0.100  m)^{2}} = 10  MPa$

Since σ is smaller than the allowable stress, a 100 × 100-mm cross section is acceptable.

(b) For the 200-kN Load. Solving again Eq. (16.11) for I, but making now $P_{cr} = 2.5(200) = 500  kN$, we have

$I = 15.588 × 10^{-6}  m^{4}$

$\frac{a^{4}}{12} = 15.588 × 10^{-6}                     a = 116.95  mm$

The value of the normal stress is

$σ = \frac{P}{A} =\frac{200  kN}{(0.11695  m)^{2}} = 14.62  MPa$

Since this value is larger than the allowable stress, the dimension obtained is not acceptable, and we must select the cross section on the basis of its resistance to compression. We write

$A =\frac{P}{σ_{all}} = \frac{200  kN}{12  MPa} = 16.67 × 10^{-3}  m^{2}$

$a^{2} = 16.67 × 10^{-3}  m^{2}                      a = 129.1  mm$

A 130 × 130-mm cross section is acceptable.